Counterexamples to Calabi conjectures on minimal hypersurfaces ...

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that yields ∑ lt=2 b2 it ≤ 1.Since ψ(x) → ∞ as ρ M (x) = dist M (q, x) → +∞, off a compact set we mayassume that√| H|(x) → = m|H|(x) ≤ ψ(x)G( √ ψ(x))where G(t) is given by (7). Otherwise, sup M |H| = +∞ and there is nothing <strong>to</strong>prove. Besides, off a compact set we also have that√1ψ(x) ≤ ψ(x)G( √ ψ(x)).Hence, from (4) we have off a compact set that∆ M ψ(x) =m∑Hess N×R l σ(ϕ(x))(e i , e i ) + 〈grad N×Rl σ(ϕ(x)), H → (x)〉i=1≤mψ(x) + m|H|(x)√≤ (m + 1) ψ(x)G( √ ψ(x)).Therefore, by Theorem 3.2 the Omori-Yau maximum principle holds on M.Define ρ: N n−l × R l → R byand u: M m → R byρ(z, y) = ρ N (z) = dist N (p, z)u(x) = ρ ◦ ϕ(x).Since ϕ(M) ⊂ B N (r) × R l , we have that u ∗ = sup M u ≤ r < ∞, Therefore, by themaximum principle there is a sequence {x k } k∈N ⊂ M m such thatHence, we have1k > ∆u(x k) =u(x k ) > u ∗ − 1/k; |gradu|(x k ) < 1/k; ∆u(x k ) < 1/k.m∑Hess N×R lρ(ϕ(x k ))(e i , e i ) + 〈grad N×Rl ρ(ϕ(x k )), H → (x k )〉 (8)i=1where {e 1 , . . .,e m } is an orthonormal basis for T xk M. Start with an orthonormalbasis {∂/∂ρ N , ∂/∂θ 2 , . . .,∂/∂θ n−l } for TN and standard coordinates {y 1 , . . .y l } forR l . Then, choose an orthonormal basis for T xk M as follows∂ ∑n−l∂e i = α i + a ij +∂ρ N ∂θ jj=26l∑t=1c it∂∂y t·
Using Theorem 3.1, a straightforward computation yieldssinceHess N×R lρ(ϕ(x k ))(e i , e i ) = Hess N ρ N (z(x k ))(π TN e i , π TN e i )=≥=∑n−la 2 ijHess N ρ N (z(x k ))(∂/∂θ j , ∂/∂θ j )j=2∑n−la 2 ijC b (r) (9)j=2(1 − α 2 i −l∑t=1c 2 it)n−l|e i | = 1 = αi 2 + ∑ l∑a 2 ij + c 2 it ,j=2t=1C b (r)where π TN denotes the orthogonal projection on<strong>to</strong> TN. Therefore,m∑Hess N×R lρ(ϕ(x k ))(e i , e i ) ≥i=1At x k , we haveand hence(m − ∑ iα 2 i − ∑ i,tgrad N×Rl ρ(ϕ(x k )) = gradu(x k ) + (grad N×Rl ρ(ϕ(x k ))) ⊥|gradu| 2 (x k ) =m∑i=1〈 ∂∂ρ N, e i 〉 = ∑ ic 2 it)C b (r). (10)α 2 i < 1/k2 . (11)Taking in<strong>to</strong> account |grad N×Rl ρ| = |grad N ρ N | = 1, from (8) and (10) we obtain1(k > m − ∑ iα 2 i − ∑ i,tc 2 it)C b (r) − m sup |H|.MIt follows using (11) that1k + C b(r)k 2(+ m sup |H| ≥ m − ∑Mi,tc 2 it)C b (r). (12)Observe now that∑c 2 it =i,tl∑ m∑c 2 it =t=1 i=1l∑|grad(y t ◦ ϕ)| 2 ≤ l,t=17
Page 1 and 2: Counterexamples <s
Page 3 and 4: 2 The resultsPart (a) of Theorem 2.
Page 5: where G is a smooth function on [0,
Page 9 and 10: m − ∑ i,tsince φ ′′b (t)

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